If f 0 - 1-f 2 - 3-f- 2 - 5 is finite then- -01 f- 2xdx is equal to

The correct option is A ZeroSolution #160; I = ∫_0^1f'(2x)dx Let 2n = t #160; #160;at n = 0 #160; t = 0 2dn = dt #160; ...

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Byju's Answer

Standard XII

Physics

Basic Differentiation Rule

If f 0 = 1,...

Question

If $f (0) = 1,f (2) = 3,f' (2) = 5$ is finite then, $1 \int 0 {f'}^{'} (2 x) dx is equal to$

Zero

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Solution

The correct option is A Zero
Solution
$I = \int_{0}^{1} f^{'} (2 x) d x$
Let $2 n = t$ at $n = 0$ $t = 0$
$2 d n = d t$ $n = 1$ $t = 2$
$∴ I = \frac{1}{2} \int_{0}^{2} f^{'} (t) d t = [f (t)]_{0}^{2} = \frac{1}{2} (f (2) - f (0))$
$= \frac{1}{2} [3 - 1] = 1$

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If f(0) = 1,f(2) = 3,f'(2) = 5 is finite then, 1∫0 f'′(2x)dxisequalto

The correct option is A ZeroSolution I=∫10f′(2x)dxLet 2n=t at n=0 t=02dn=dt n=1 t=2 ∴I=12∫20f′(t)dt=[f(t)]20=12(f(2)−f(0))=12[3−1]=1

If $f (0) = 1,f (2) = 3,f' (2) = 5$ is finite then, $1 \int 0 {f'}^{'} (2 x) dx is equal to$

The correct option is A Zero
Solution
$I = \int_{0}^{1} f^{'} (2 x) d x$
Let $2 n = t$ at $n = 0$ $t = 0$
$2 d n = d t$ $n = 1$ $t = 2$
$∴ I = \frac{1}{2} \int_{0}^{2} f^{'} (t) d t = [f (t)]_{0}^{2} = \frac{1}{2} (f (2) - f (0))$
$= \frac{1}{2} [3 - 1] = 1$